Chemical explosive reaction


Rapidity of reaction[22] distinguishes the explosive reaction from an ordinary combustion reaction by the great speed with which it takes place. Unless the reaction occurs rapidly, the thermally expanded gases will be dissipated in the medium, and there will be no explosion. Consider a wood or coal fire. As the fire burns, there is the evolution of heat and the formation of gases, but neither is liberated rapidly enough to cause an explosion. This can be likened to the difference between the energy discharge of a battery, which is slow, and that of a flash capacitor like that in a camera flash, which releases its energy all at once.

There are many chemical reactions that will release energy[20] . These are known as exothermic reactions. If the reaction proceeds slowly, the released energy will be dissipated and there will be few noticeable effects other than an increase in temperature. On the other hand, if the reaction proceeds very rapidly, then the energy will not be dissipated. Thus, a great quantity of energy can be deposited into a relatively small volume, then manifest itself by a rapid expansion of hot gases, which in turn can create a shock wave or propel fragments outwards at high speed. Chemical explosions may be distinguished from other exothermic reactions by the extreme rapidity of their reactions. In addition to the violent release of energy, chemical explosions must provide a means to transfer the energy into mechanical work. This is accomplished by expanding product gases from the reaction. If no gases are produced, then the energy will remain in the products as heat.

Most chemical explosions involve a limited set of simple reactions, all of which involve oxidation (reaction with oxygen). A relatively easy way to balance chemical explosive equations is to assume that the following partial reactions take place to their maximum extent (meaning one of the reactants is totally consumed) and in order of precedence:

Priority
Reaction (to completion)
1
Metal + O → Metallic Oxide (ex: ZnO or PbO)
2
C + O → CO (gas)
3
2H + O → H2O (gas)
4
CO + O → CO2 (gas) (The CO comes from reaction (2))
5
Excess O, H, NO2, N2, H2 (gases)

The characteristics of an explosive[5] [17] are quantities, often but not always experimentally determinable with precision, which enable to bring into focus the effects it is capable to exercise. The most significant are: a) oxygen balance b) detonation velocity c) specific volume d) heat of explosion e) temperature of explosion f) specific pressure.

a) Oxygen balance
The reaction course in an explosive compound or mixture can vary with the variation of certain conditions, for example of the intensity of the initial impulse, or of the reached temperature and pression, which, as is common knowledge, have considerable influence on the chemical equilibria between reaction products. Neverthless it is possible to know the equation of decomposition with good certainty if the esplosive contains in itself oxygen enough to the complete oxidation of his own molecule or in excess compared to it: so the balance of oxygen is called, respectively, zero or positive.
Carbon turns into CO2, Hydrogen in H2O and Nitrogen in N2, while the possible metals originate the corresponding oxides.
An example of zero balance is given by Nitroglycol:

C2H4(NO3)2 → 2CO2 + 2H2O + N2

an example of positive balance is given by Nitroglycerine:

4C3H5(NO3)2 → 12CO2 + 10H2O + 6N2 + O2

When, instead, oxygen is insufficient for a complete oxidation it distribute itself between Carbon and Hydrogen in accordance with the complex dissociation equilibria of CO2 and H2O. It is therefore necessary to set the calculations on the base of the knowledge of these equilibria at the explosion temperature.

b) Detonation velocity
It can be experimentally determined with good precision: one of the most functional and most used method still remains the old “Dautriche method”.
A represents the cartridge of the explosive under consideration in which, in the holes B and C, are inserted two detonators locked up at the edges of a detonating fuse with well known detonation velocity, D. The detonating fuse is fixed in his central part with copper wires on a lead plate F so that the point P, marked on the plate, coincide perfectly with the half of the fuse length.
Since BP and CP distances are equal, if the two detonators B and C were primed at the same istant the two explosive waves would go back to the edges of the fuse in order to reach at the same istant P, where this frontal meeting would leave a sign engraved on the lead plate.
In the practical execution, by means of detonator E, the explosive under consideration is primed and this explosive give start in turn to detonator in position B and subsequently to detonator in C.
The two explosive waves meet no more in P but in a new point P'.

Dautriche method for the determination of detonation velocity:
A=explosive
B, C=holes for the detonators
D=detonating fuse
E=detonators
F=lead plate

The time requested by the detonation of the explosive under consideration to spread from B to C will be equal to the difference of the detonation course in the two tracts of detonating fuse, i.e. 2PP'. So:

Dautriche method has the inconvenient that is an indirect method: the measurement is made in comparison with an explosive having known velocity. To measure the sample it will be necessary to resort to direct methods.
Mettegang mehod is direct: it is based on the measurement of the traces marked on the blackened surface of a rotating drum when the detonation interrupts two or more electric circuits and causes the shoot out of the corresponding sparks between a point of Platinum and the drum which is rotating known speed.
There are other direct methods based on different principles, for example electrophotograpic or based on the drop of potential of a condenser when the detonation interrupt subsequently two circuits.

c) Specific volume
Specific or normal volume (Vk) is the volume of gas that 1 kg of explosive can generate at a temperature of 0°C and at a pressure of 760 mmHg. In can be determinated experimentally with the manometric bomb or, much easily, with a theoretical calculation provided as the stechiometry of the reaction is known.
So, since 4 mole (980 g) of nitroglycerine generate 29 mole of gas:




Because of the high thermal expansion of gases, the volume at the explosion temperature, 4000°C for nitroglycerine, becomes considerably higher.

d) Heat of explosion[6]
The heat of explosion is the energy, at constant volume express in kcal, released by 1kg of explosive.
It can be determined experimentally with the calorimetric bomb, or calculating the difference between the sum of formation enthalpies of the products of explosion and the one of the components of the explosive.So, for nitroglycerine:

4C3H5(NO3)3
12CO2 + 10H2O + 6N2 + O2
4 x 88,4
12 x 94,4 + 10 x 57,8

That is 4 mole of nitroglycerine (908g) generate 1357 kcal at constant pressure.
In a reaction in which n mole of gas are formed, the heat at constant volume is different from the one at constant pressure because of the work done during the Expansion at constant pressure



For nitroglycerine:



that corresponds, for 1 kg, to 1513 kcal.
Another quantity used for practical intents is the potential, that is the mechanical equivalent of the heat of expansion. It represents, but only theoretically, the work an explosive is able to perform.



For nitroglycerine:




e) Temperature of explosion
The direct determination of the temperature of explosion is somewhat problematic because it must be done in the short instants when the maximum value is reached. However this temperature can be calculated by means of the heat of explosion at constant volume and of the specific heats of reaction products by using the equation:





But specific heat aren’t constant: they vary with pressure and also with temperature. If the influence of pressure is ignored, the variation of specific heats can be represented by the equation:




If we integrate the firs equation between ambient temperature (t0) and temperature of explosion (t), ignoring the value t0 compared to t and limiting the development of the equation at the second term, we have:




Because we have the specific heats of N products we must take in consideration the summations:





We obtain, by the positive solution:





aand b values have been determined experimentally by several authors and the most important are those showed in the table below:

In the case of nitroglycerine, by using Kast’s coefficients, since from the explosion of 4 mole of nitroglycerine we obtain 12 mole of CO2, 10 mole of H2O, 6 mole of NO2 and 1 mole of O2, we obtain:









So, for 4 mole of nitroglycerine andare 181,60 and 0,04725 cal/mole which, compared to 1kg, give respectively 199,91 and 0,05201.





The value of 3803°C (4080°K) calculated in this way tends to be lower the the ones generally indicated by authors, who report different data depending of used coefficients of methods. For the calculation of the other characteristics we have chosen, here, the more reliable value of 4000°C.

f) Specific pressure
The pressure that 1kg of explosive is able to release is derivable form the ideal gases equation:





Where T is the absolute temperature of explosion and P0 and Vk are, respectively, normal pressure and specific volume.
If we suppose that the explosion happens within 1litre volume, we have:





The f quantity, which is a constant characteristic for every explosive, is improperly called “specific pressure” because, in point of fact, it has the dimensions of an energy and it’s expressed in litres x atmospheres.





Kast suggested that the exploding power could be expressed with the “brisance”, defining it with the formula:




Where d is the density. For nitroglycerine: